3.4.67 \(\int \frac {\sqrt {a+b x} (A+B x)}{x} \, dx\)

Optimal. Leaf size=54 \[ 2 A \sqrt {a+b x}-2 \sqrt {a} A \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )+\frac {2 B (a+b x)^{3/2}}{3 b} \]

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Rubi [A]  time = 0.02, antiderivative size = 54, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {80, 50, 63, 208} \begin {gather*} 2 A \sqrt {a+b x}-2 \sqrt {a} A \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )+\frac {2 B (a+b x)^{3/2}}{3 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(Sqrt[a + b*x]*(A + B*x))/x,x]

[Out]

2*A*Sqrt[a + b*x] + (2*B*(a + b*x)^(3/2))/(3*b) - 2*Sqrt[a]*A*ArcTanh[Sqrt[a + b*x]/Sqrt[a]]

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 80

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(c + d*x)
^(n + 1)*(e + f*x)^(p + 1))/(d*f*(n + p + 2)), x] + Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(
d*f*(n + p + 2)), Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2,
0]

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rubi steps

\begin {align*} \int \frac {\sqrt {a+b x} (A+B x)}{x} \, dx &=\frac {2 B (a+b x)^{3/2}}{3 b}+A \int \frac {\sqrt {a+b x}}{x} \, dx\\ &=2 A \sqrt {a+b x}+\frac {2 B (a+b x)^{3/2}}{3 b}+(a A) \int \frac {1}{x \sqrt {a+b x}} \, dx\\ &=2 A \sqrt {a+b x}+\frac {2 B (a+b x)^{3/2}}{3 b}+\frac {(2 a A) \operatorname {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b x}\right )}{b}\\ &=2 A \sqrt {a+b x}+\frac {2 B (a+b x)^{3/2}}{3 b}-2 \sqrt {a} A \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )\\ \end {align*}

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Mathematica [A]  time = 0.07, size = 55, normalized size = 1.02 \begin {gather*} A \left (2 \sqrt {a+b x}-2 \sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )\right )+\frac {2 B (a+b x)^{3/2}}{3 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(Sqrt[a + b*x]*(A + B*x))/x,x]

[Out]

(2*B*(a + b*x)^(3/2))/(3*b) + A*(2*Sqrt[a + b*x] - 2*Sqrt[a]*ArcTanh[Sqrt[a + b*x]/Sqrt[a]])

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IntegrateAlgebraic [A]  time = 0.03, size = 57, normalized size = 1.06 \begin {gather*} \frac {2 \left (3 A b \sqrt {a+b x}+B (a+b x)^{3/2}\right )}{3 b}-2 \sqrt {a} A \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(Sqrt[a + b*x]*(A + B*x))/x,x]

[Out]

(2*(3*A*b*Sqrt[a + b*x] + B*(a + b*x)^(3/2)))/(3*b) - 2*Sqrt[a]*A*ArcTanh[Sqrt[a + b*x]/Sqrt[a]]

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fricas [A]  time = 1.20, size = 111, normalized size = 2.06 \begin {gather*} \left [\frac {3 \, A \sqrt {a} b \log \left (\frac {b x - 2 \, \sqrt {b x + a} \sqrt {a} + 2 \, a}{x}\right ) + 2 \, {\left (B b x + B a + 3 \, A b\right )} \sqrt {b x + a}}{3 \, b}, \frac {2 \, {\left (3 \, A \sqrt {-a} b \arctan \left (\frac {\sqrt {b x + a} \sqrt {-a}}{a}\right ) + {\left (B b x + B a + 3 \, A b\right )} \sqrt {b x + a}\right )}}{3 \, b}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(b*x+a)^(1/2)/x,x, algorithm="fricas")

[Out]

[1/3*(3*A*sqrt(a)*b*log((b*x - 2*sqrt(b*x + a)*sqrt(a) + 2*a)/x) + 2*(B*b*x + B*a + 3*A*b)*sqrt(b*x + a))/b, 2
/3*(3*A*sqrt(-a)*b*arctan(sqrt(b*x + a)*sqrt(-a)/a) + (B*b*x + B*a + 3*A*b)*sqrt(b*x + a))/b]

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giac [A]  time = 1.34, size = 55, normalized size = 1.02 \begin {gather*} \frac {2 \, A a \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{\sqrt {-a}} + \frac {2 \, {\left ({\left (b x + a\right )}^{\frac {3}{2}} B b^{2} + 3 \, \sqrt {b x + a} A b^{3}\right )}}{3 \, b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(b*x+a)^(1/2)/x,x, algorithm="giac")

[Out]

2*A*a*arctan(sqrt(b*x + a)/sqrt(-a))/sqrt(-a) + 2/3*((b*x + a)^(3/2)*B*b^2 + 3*sqrt(b*x + a)*A*b^3)/b^3

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maple [A]  time = 0.01, size = 46, normalized size = 0.85 \begin {gather*} \frac {-2 A \sqrt {a}\, b \arctanh \left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )+2 \sqrt {b x +a}\, A b +\frac {2 \left (b x +a \right )^{\frac {3}{2}} B}{3}}{b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*(b*x+a)^(1/2)/x,x)

[Out]

2/b*(1/3*B*(b*x+a)^(3/2)+A*b*(b*x+a)^(1/2)-A*a^(1/2)*b*arctanh((b*x+a)^(1/2)/a^(1/2)))

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maxima [A]  time = 1.96, size = 60, normalized size = 1.11 \begin {gather*} A \sqrt {a} \log \left (\frac {\sqrt {b x + a} - \sqrt {a}}{\sqrt {b x + a} + \sqrt {a}}\right ) + \frac {2 \, {\left ({\left (b x + a\right )}^{\frac {3}{2}} B + 3 \, \sqrt {b x + a} A b\right )}}{3 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(b*x+a)^(1/2)/x,x, algorithm="maxima")

[Out]

A*sqrt(a)*log((sqrt(b*x + a) - sqrt(a))/(sqrt(b*x + a) + sqrt(a))) + 2/3*((b*x + a)^(3/2)*B + 3*sqrt(b*x + a)*
A*b)/b

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mupad [B]  time = 0.07, size = 45, normalized size = 0.83 \begin {gather*} 2\,A\,\sqrt {a+b\,x}+\frac {2\,B\,{\left (a+b\,x\right )}^{3/2}}{3\,b}+A\,\sqrt {a}\,\mathrm {atan}\left (\frac {\sqrt {a+b\,x}\,1{}\mathrm {i}}{\sqrt {a}}\right )\,2{}\mathrm {i} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*x)*(a + b*x)^(1/2))/x,x)

[Out]

2*A*(a + b*x)^(1/2) + A*a^(1/2)*atan(((a + b*x)^(1/2)*1i)/a^(1/2))*2i + (2*B*(a + b*x)^(3/2))/(3*b)

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sympy [A]  time = 5.78, size = 54, normalized size = 1.00 \begin {gather*} \frac {2 A a \operatorname {atan}{\left (\frac {\sqrt {a + b x}}{\sqrt {- a}} \right )}}{\sqrt {- a}} + 2 A \sqrt {a + b x} + \frac {2 B \left (a + b x\right )^{\frac {3}{2}}}{3 b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(b*x+a)**(1/2)/x,x)

[Out]

2*A*a*atan(sqrt(a + b*x)/sqrt(-a))/sqrt(-a) + 2*A*sqrt(a + b*x) + 2*B*(a + b*x)**(3/2)/(3*b)

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